Define $T:P(R)->P(R)$ by $Tp=p'.$ Find all eigen values and eigenvectors of T Define $T:P(R)->P(R)$ by $Tp=p'.$ Find all eigen values and eigenvectors of T
I sort of have a solution but I am trying to make sense of it.
My attempt:
Suppose $\lambda$ is an eigenvalue of T with eigenvector $v\in V$. Then by definition, $v\neq 0$ and $Tv=\lambda v$. Now
$v'=Tv=\lambda v$. We can observe that if $\lambda \neq 0$ Then
$deg(v')<deg(v)$
$deg(v')<deg(\lambda v)$
Since the degree of a derivative is less than the degree of the original variable. *Now it says that we get a contradiction. My first question is... Why do we get a contradiction?**
Continuing the proof...
If $\lambda =0$ then
$v'=Tv=0v$ which implies $v'=0$. Since $v'=0$ $v$ must be a constant since the derivative of a constant is equal to zero.*Now it says....Hence the only eigenvalue of T is zero with nonzero constant polynomials as eigenvectors...My second question... why is this true?**
 A: Suppose $\lambda \neq 0$ and $v=a_n x^n +\cdots + a_1 x + a_0$ is a polynomial of degree $n > 0$ (i.e. $a_n\neq 0$) for which $v' = \lambda v$.
Then $$v'= 0 x^{n} + a_n n x^{n-1} + \cdots + 2a_2 x + a_1 = \lambda v = \lambda a_n x^n + \lambda a_{n-1} x^{n-1} + \cdots + \lambda a_0.$$ The zero leading coefficient was added for emphasis.
Thus by comparing coefficients we have $\lambda a_{i-1} = i a_i$ for each $i=1,...,n$, and $a_n = 0$ ($v'$ has degree less than $n$), since two polynomials agree iff their coefficients agree. This contradicts our assumption of the degree of $v$ being $n$, since $a_n$ is supposed to be nonzero.
Thus the degree of $v$ must be $0$, and $v$ is a constant. Now we know that for any constant function $v$, $v' =0 = 0 \cdot v$. Thus, the constant polynomials are eigenfunctions (eigenvectors) with eigenvalue $0$.
A: Let $\mathcal{E}$ the standard basis for $P(R)$. Then the matrix of $T$ respect to $\mathcal{E}$ is upper triangular with all zeros in the main diagonal. So the unique eigenvalue is $0$ and clearly the matrix is not diagonalizable. The eigenspace is given by the constant functions, since the eigenspace relative to $0$ coincide with the kernel of $T$.
